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\noindent {\bf Question}
\noindent For each of the following functions described below,
determine whether there is a solution to the given equation in the
specified set.
\begin{enumerate}
\item $g'(a)=0=g'(b)$, where $a**0$;
\end{enumerate}
\medskip
\noindent {\bf Answer}
\noindent (In these problems, I've stopped explicitly checking the
continuity and diffentiability hypotheses of the intermediate
value property and of Rolle's theorem and the mean value theorem,
because they have been checked so many times already and since
they hold true for all the functions in this exercise.)
\begin{enumerate}
\item using the general mantra that two solutions to $g(x) =0$ yield
one solution to $g'(x) =0$ via Rolle's theorem, let's see if we
can find three solutions to $g(x) =0$ for $g(x) = x^3 - 12\pi x^2
+ 44\pi^2 x - 48\pi^3 + \cos(x) - 1$. Factoring, we see that
$g(x) = (x-2\pi)(x-4\pi)(x-6\pi) + \cos(x) -1$, and so $g(2\pi) =
g(4\pi) = g(6\pi) =0$. By Rolle's theorem, there then exists $a$
in $(2\pi, 4\pi)$ and $b$ in $(4\pi, 6\pi)$ so that $g'(a) =g'(b)
=0$, as desired. (Also, note that the mixture of polynomial and
trigonometric functions makes it unlikely that we would find
solutions to $g'(x) =0$ by direct calculation.)
\item a still slightly different method: calculating, we see that
$f'(x)=4x^3 - \pi^3 -\cos(x)$, and that $f'(-10) = -4000 -\pi^3
-\cos(-1000) <0$ and that $f'(10) = 4000 -\pi^3 -\cos(1000) >0$.
Since $f$ is continuous on ${\bf R}$, it is certainly continuous
on the interval $[-10, 10]$, and so by the intermediate value
property, there is some $a$ in $(-10,10)$ at which $f'(a) =0$.
\item label the points at which $g$ vanishes as $a_1 < a_2 <\cdots <
a_n$. For each consecutive pair $a_k$, $a_{k+1}$, Rolle's theorem
yields that there exists a point $b_k$ between $a_k$ and $a_{k+1}$
at which $g'(b_k) =0$. This yields $k-1$ points $b_1,\ldots,
b_{k-1}$ at which the derivative $g'(x)$ vanishes, as desired.
\item let $h(x) = x^3 +px +q$. Suppose that $h$ has two real roots;
by Rolle's theorem, there is then a number $c$ between these roots
at which $h'(c) =0$. However, calculating directly we see that
$h'(x) = 3x^2 +p \ge p >0$ for all $x\in {\bf R}$, and so there
are no solutions to $h'(x) =0$. Hence, there can be at most one
root of $h$.
\medskip
\noindent To see that there is a root, we note that since $h$ has
odd degree (and since the coefficient of the highest degree term
is positive), we have that $\lim_{x\rightarrow\infty} h(x)
=\infty$ and $\lim_{x\rightarrow -\infty} h(x) =-\infty$. Hence,
we can find a point $a$ at which $h(a) >0$ and a point $b$ at
which $h(b) <0$, and the intermediate value property then implies
that there is a point between $a$ and $b$ at which $h(x) =0$.
\end{enumerate}
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